@Starwar Clone
–
Tomorrow, I rewiewed this, and if it's not solved. I'll try to solve it, I think I'm not going to have a lot problemms.. we'll see...
–
Guillermo Templado
·
2 months, 2 weeks ago

Log in to reply

@Starwar Clone
–
I agree with Pi Hang Goh, you can also make the change \(t = x\Sh(u)\) solve , the integral, and later derivate
–
Guillermo Templado
·
2 months, 2 weeks ago

@Pi Han Goh
–
yeah but i am still not able to get the required answer will you please post the solution to this question...
–
Starwar Clone
·
2 months, 2 weeks ago

Log in to reply

@Starwar Clone
–
The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.
–
Pi Han Goh
·
2 months, 2 weeks ago

Log in to reply

@Pi Han Goh
–
i am getting the answer \[ \dfrac{3x^2}{\sqrt{x^2 + x^12 }} \]
it is x^12 .. in the denominator
–
Starwar Clone
·
2 months, 2 weeks ago

Log in to reply

@Starwar Clone
–
Split the integral into \( \int_0^{x^3} - \int_0^{t^2} \). For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you?
–
Pi Han Goh
·
2 months, 2 weeks ago

@Starwar Clone
–
Make the change I have said to you... It must solve it quickly.. or use Fundamental theorem of calculus
–
Guillermo Templado
·
2 months, 2 weeks ago

@Starwar Clone
–
It's not necessary, but if you make the change \(t = x \cdot \text{ Sh u} \) with \(x\) constant, and use \(1 + Sh^2(u) = Ch^2(u)\) and \(dt = x \cdot \text{Ch u} \space du\), the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise....
–
Guillermo Templado
·
2 months, 2 weeks ago

@Starwar Clone
–
I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also if \(x >0\), \(x = 0\) or \(x < 0\) to concrete the ending. This question is giving me questions and suggestions... For example, does \(\int_0^{t} \space dt\) make some sense?If the answer is yes, Is it \(t\)?why?... I have another proof making the change \(u = \sqrt{x^2 + t^2} + t\)... Even we can think first this question without the derivative \(\frac{d}{dx}\), I mean given \(x \) possible concrete constants, \(x =1\), \(x =2\), \(x = 100\)... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare...
–
Guillermo Templado
·
2 months, 2 weeks ago

## Comments

Sort by:

TopNewest@Pi Han Goh @Guillermo Templado help will be welcomed . – Starwar Clone · 2 months, 2 weeks ago

Log in to reply

– Guillermo Templado · 2 months, 2 weeks ago

Tomorrow, I rewiewed this, and if it's not solved. I'll try to solve it, I think I'm not going to have a lot problemms.. we'll see...Log in to reply

– Guillermo Templado · 2 months, 2 weeks ago

I agree with Pi Hang Goh, you can also make the change \(t = x\Sh(u)\) solve , the integral, and later derivateLog in to reply

– Guillermo Templado · 2 months, 2 weeks ago

I mean \( t = x \cdot \text{Sh u}\). I have some problems with my comments..Log in to reply

Hint: Fundamental Theorem of Calculus. – Pi Han Goh · 3 months agoLog in to reply

– Starwar Clone · 2 months, 2 weeks ago

yeah but i am still not able to get the required answer will you please post the solution to this question...Log in to reply

– Pi Han Goh · 2 months, 2 weeks ago

The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.Log in to reply

– Starwar Clone · 2 months, 2 weeks ago

i am getting the answer \[ \dfrac{3x^2}{\sqrt{x^2 + x^12 }} \] it is x^12 .. in the denominatorLog in to reply

– Pi Han Goh · 2 months, 2 weeks ago

Split the integral into \( \int_0^{x^3} - \int_0^{t^2} \). For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you?Log in to reply

– Starwar Clone · 2 months, 2 weeks ago

Yeah that is enoughLog in to reply

– Starwar Clone · 2 months, 2 weeks ago

Is the answer correct?Log in to reply

– Guillermo Templado · 2 months, 2 weeks ago

Make the change I have said to you... It must solve it quickly.. or use Fundamental theorem of calculusLog in to reply

– Starwar Clone · 2 months, 2 weeks ago

can you provide the final answer , i am really stuck on this ?Log in to reply

– Guillermo Templado · 2 months, 2 weeks ago

you have to develop the steps...Log in to reply

– Starwar Clone · 2 months, 2 weeks ago

do you mean that i really need to integrate the above expression first?Log in to reply

– Guillermo Templado · 2 months, 2 weeks ago

It's not necessary, but if you make the change \(t = x \cdot \text{ Sh u} \) with \(x\) constant, and use \(1 + Sh^2(u) = Ch^2(u)\) and \(dt = x \cdot \text{Ch u} \space du\), the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise....Log in to reply

– Starwar Clone · 2 months, 2 weeks ago

thanks man , appreciate the helpLog in to reply

Proof 1 (attempt 1).-\[\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} \left( \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt - \int_{0}^{t^2} \dfrac 1{\sqrt{x^2 + t^2}} \, dt \right) =\] Let's call \(g(x) = x^3\) and \(f(x) = \int_{0}^{x} \dfrac 1{\sqrt{x^2 + t^2}} \, dt\) Then, \[\dfrac d{dx} \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} f(g(x)) = \dfrac d{dx} f(x^3) = 3x^2 \cdot ...\] Sorry, I'm getting a knot for myself...

Proof 2.-\[\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \space (*)\] Let's make the change \( t = x \cdot \text{ Sh u}\) , \(x\) constant, \(dt = x \cdot \text{Ch u} \space du\) \[ (*) = \dfrac d{dx} \int_{Sh^{-1}(x \cdot \text{ Sh}^{2}u)}^{Sh^{-1} x^2} \, \pm \space du =...\] To be continued – Guillermo Templado · 2 months, 2 weeks ago

Log in to reply

– Starwar Clone · 2 months, 2 weeks ago

Hmm it's getting betterLog in to reply

– Guillermo Templado · 2 months, 2 weeks ago

I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also if \(x >0\), \(x = 0\) or \(x < 0\) to concrete the ending. This question is giving me questions and suggestions... For example, does \(\int_0^{t} \space dt\) make some sense?If the answer is yes, Is it \(t\)?why?... I have another proof making the change \(u = \sqrt{x^2 + t^2} + t\)... Even we can think first this question without the derivative \(\frac{d}{dx}\), I mean given \(x \) possible concrete constants, \(x =1\), \(x =2\), \(x = 100\)... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare...Log in to reply